3.594 \(\int \frac{A+B x^2}{x^3 (a+b x^2)^{5/2}} \, dx\)
Optimal. Leaf size=113 \[ -\frac{5 A b-2 a B}{2 a^3 \sqrt{a+b x^2}}-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}} \]
[Out]
-(5*A*b - 2*a*B)/(6*a^2*(a + b*x^2)^(3/2)) - A/(2*a*x^2*(a + b*x^2)^(3/2)) - (5*A*b - 2*a*B)/(2*a^3*Sqrt[a + b
*x^2]) + ((5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(7/2))
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Rubi [A] time = 0.0865613, antiderivative size = 113, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used =
{446, 78, 51, 63, 208} \[ -\frac{5 A b-2 a B}{2 a^3 \sqrt{a+b x^2}}-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*x^2)/(x^3*(a + b*x^2)^(5/2)),x]
[Out]
-(5*A*b - 2*a*B)/(6*a^2*(a + b*x^2)^(3/2)) - A/(2*a*x^2*(a + b*x^2)^(3/2)) - (5*A*b - 2*a*B)/(2*a^3*Sqrt[a + b
*x^2]) + ((5*A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(7/2))
Rule 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 78
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ
[p, n]))))
Rule 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{A+B x^2}{x^3 \left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{x^2 (a+b x)^{5/2}} \, dx,x,x^2\right )\\ &=-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}}+\frac{\left (-\frac{5 A b}{2}+a B\right ) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{5/2}} \, dx,x,x^2\right )}{2 a}\\ &=-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac{(5 A b-2 a B) \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^{3/2}} \, dx,x,x^2\right )}{4 a^2}\\ &=-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 A b-2 a B}{2 a^3 \sqrt{a+b x^2}}-\frac{(5 A b-2 a B) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a^3}\\ &=-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 A b-2 a B}{2 a^3 \sqrt{a+b x^2}}-\frac{(5 A b-2 a B) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 a^3 b}\\ &=-\frac{5 A b-2 a B}{6 a^2 \left (a+b x^2\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^2\right )^{3/2}}-\frac{5 A b-2 a B}{2 a^3 \sqrt{a+b x^2}}+\frac{(5 A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{7/2}}\\ \end{align*}
Mathematica [C] time = 0.0210112, size = 57, normalized size = 0.5 \[ \frac{x^2 (2 a B-5 A b) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b x^2}{a}+1\right )-3 a A}{6 a^2 x^2 \left (a+b x^2\right )^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*x^2)/(x^3*(a + b*x^2)^(5/2)),x]
[Out]
(-3*a*A + (-5*A*b + 2*a*B)*x^2*Hypergeometric2F1[-3/2, 1, -1/2, 1 + (b*x^2)/a])/(6*a^2*x^2*(a + b*x^2)^(3/2))
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Maple [A] time = 0.01, size = 140, normalized size = 1.2 \begin{align*}{\frac{B}{3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{B}{{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{B\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{A}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,Ab}{6\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,Ab}{2\,{a}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{5\,Ab}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{7}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x)
[Out]
1/3*B/a/(b*x^2+a)^(3/2)+B/a^2/(b*x^2+a)^(1/2)-B/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-1/2*A/a/x^2/(b*x
^2+a)^(3/2)-5/6*A*b/a^2/(b*x^2+a)^(3/2)-5/2*A*b/a^3/(b*x^2+a)^(1/2)+5/2*A*b/a^(7/2)*ln((2*a+2*a^(1/2)*(b*x^2+a
)^(1/2))/x)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.70717, size = 756, normalized size = 6.69 \begin{align*} \left [-\frac{3 \,{\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} +{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{a} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (3 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \,{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{12 \,{\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, \frac{3 \,{\left ({\left (2 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 2 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} +{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \,{\left (2 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{4} - 3 \, A a^{3} + 4 \,{\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="fricas")
[Out]
[-1/12*(3*((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2*B*a^2*b - 5*A*a*b^2)*x^4 + (2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(a)*log(
-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*(2*B*a^2*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*
A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), 1/6*(3*((2*B*a*b^2 - 5*A*b^3)*x^6 + 2*(2
*B*a^2*b - 5*A*a*b^2)*x^4 + (2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(2*B*a^2
*b - 5*A*a*b^2)*x^4 - 3*A*a^3 + 4*(2*B*a^3 - 5*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6
*x^2)]
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Sympy [B] time = 49.8156, size = 1608, normalized size = 14.23 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x**2+A)/x**3/(b*x**2+a)**(5/2),x)
[Out]
A*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2
)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2
*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*
a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70*a**15*b**2*x**4*sqrt(1 + b*x**2
/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b**
2*x**4*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**
8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b*
*2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*
x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14*b**3*x**6*log(sqrt(1 + b*x
**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a
**13*b**4*x**8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*
b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(
35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8)) + B*(8*a**7*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 +
18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 3*a**7*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 1
8*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**7*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2
)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 14*a**6*b*x**2*sqrt(1 + b*x**2/a)/(6*a**(19/2) +
18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 9*a**6*b*x**2*log(b*x**2/a)/(6*a**(19/
2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 18*a**6*b*x**2*log(sqrt(1 + b*x**
2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) + 6*a**5*b**2*x
**4*sqrt(1 + b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) +
9*a**5*b**2*x**4*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*
x**6) - 18*a**5*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2)*b**2*x
**4 + 6*a**(13/2)*b**3*x**6) + 3*a**4*b**3*x**6*log(b*x**2/a)/(6*a**(19/2) + 18*a**(17/2)*b*x**2 + 18*a**(15/2
)*b**2*x**4 + 6*a**(13/2)*b**3*x**6) - 6*a**4*b**3*x**6*log(sqrt(1 + b*x**2/a) + 1)/(6*a**(19/2) + 18*a**(17/2
)*b*x**2 + 18*a**(15/2)*b**2*x**4 + 6*a**(13/2)*b**3*x**6))
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Giac [A] time = 1.15418, size = 136, normalized size = 1.2 \begin{align*} \frac{{\left (2 \, B a - 5 \, A b\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{2 \, \sqrt{-a} a^{3}} + \frac{3 \,{\left (b x^{2} + a\right )} B a + B a^{2} - 6 \,{\left (b x^{2} + a\right )} A b - A a b}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}} - \frac{\sqrt{b x^{2} + a} A}{2 \, a^{3} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x^2+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="giac")
[Out]
1/2*(2*B*a - 5*A*b)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 1/3*(3*(b*x^2 + a)*B*a + B*a^2 - 6*(b*x^
2 + a)*A*b - A*a*b)/((b*x^2 + a)^(3/2)*a^3) - 1/2*sqrt(b*x^2 + a)*A/(a^3*x^2)